CBSE 2025, Set 2

Question

Two dice are thrown. Defined are the following two events A and B:
$A=\{(x,y):x+y=9\}$, $B=\{(x,y):x
e3\}$, where $(x, y)$ denote a point in the sample space.
Check if events A and B are independent or mutually exclusive.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Define the sample space and events.** When two dice are thrown, the total number of outcomes in sample space $S$ is $n(S) = 6 	imes 6 = 36$. Event A: Sum of the numbers is 9. $A = \{(3, 6), (4, 5), (5, 4), (6, 3)\}$ $n(A) = 4 \implies P(A) = \frac{4}{36} = \frac{1}{9}$ Event B: The first die is NOT 3. There are 6 outcomes where the first die is 3: $\{(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\}$. Therefore, $n(B) = 36 - 6 = 30$. $P(B) = \frac{30}{36} = \frac{5}{6}$ **Step 2: Find the intersection.** Event $A \cap B$ means the sum is 9 AND the first die is not 3. From set $A$, removing the outcome that starts with 3 gives: $A \cap B = \{(4, 5), (5, 4), (6, 3)\}$ $n(A \cap B) = 3 \implies P(A \cap B) = \frac{3}{36} = \frac{1}{12}$ **Step 3: Check for Mutually Exclusive.** Events are mutually exclusive if $A \cap B = \emptyset$ (i.e., $P(A \cap B) = 0$). Since $P(A \cap B) = \frac{1}{12} 
e 0$, events A and B are **NOT mutually exclusive**. **Step 4: Check for Independence.** Events are independent if $P(A \cap B) = P(A) \cdot P(B)$. Calculate the product: $P(A) \cdot P(B) = \left(\frac{1}{9}ight) 	imes \left(\frac{5}{6}ight) = \frac{5}{54}$ Compare with intersection: $\frac{1}{12} 
e \frac{5}{54}$ Since $P(A \cap B) 
e P(A) \cdot P(B)$, events A and B are **NOT independent**.

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